Finding the limit of a complex function
I recommend to use the L'Hopital rule: $$ \lim_{z \to -3i} \frac{z^3-27i}{z+3i}=\lim_{z \to -3i} \frac{(z^3-27i)^{'}}{(z+3i)^{'}}=\lim_{z \to -3i} \frac{3z^{2}}{1}=-27. $$
$$\lim_{z \to -3i} \frac{z^3-27i}{z+3i}=\lim_{z \to -3i} \frac{(z+3i)(z^2-3 i z-9 )}{z+3i}$$
$-27i = 3^3\cdot i^3 = (3i)^3$
Now you have $z^3-27i=z^3+(3i)^3= (z+3i)(z^2+(3i)^2-3zi)$
Can you solve now?